Question

Ratio of the area cut off a parabola by any double ordinate is that of the corresponding rectangle contained by that double ordinate and its distance from the vertex is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. none of these

Answer 1

The correct option is C $\frac{2}{3}$

Let $y^2=4ax$ be a parabola and let $x=b$ be a double ordinate. Then,

$A_1=$ Area enclosed by the parabola $y^2=4ax$ and the double ordinate $x=b$

$=2{\int }_0^{b}ydx=2{\int }_0^b\sqrt{4ax}dx=4\sqrt{a}{\int }_0^b\sqrt{x^3}dx$

$=4\sqrt{a}{\left[\frac{2}{3}{x}^{3/2}\right]}_0^b=4\sqrt{a}\times \frac{2}{3}{b}^{3/2}=\frac{8}{3}{a}^{1/2}{b}^{3/2}$

And, $A_2=$ Area of the rectangle $ABCD$

$=AB\times AD=2\sqrt{4ab}\times b=4a^{1/2}{b}^{3/2}$

$\therefore A_1:A_2=\frac{8}{3}{a}^{1/2}{b}^{3/2}:4a^{1/2}{b}^{3/2}=\frac{2}{3}:1=2:3$